A Hilbert space A Hilbert space H such that on B(H)
(i) the involution isn't continuous with respect to the strong operator topology;
(ii) the weak operator topology and the strong operator topology are different;
(iii) the operator norm is not continuous with respect to the strong operator topology and the weak operator topology;
(iv) the weak operator topology and the strong operator topology aren't metrizable;
(v) the operation multiplication is continuous in neither weak nor strong operator topology.

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Let H = l2 and (en) be the standard orthonormal basis for H ( note that for all x Î H, x = ån = 1¥ < x,en > en). Set Tn = e1[`(Ä)] en. Then Tn* = en[`(Ä)] e1
(i) limn ||Tnx || = limn || < x,en > e1 || = limn | < x,en > | = 0. So Tn ® 0 in the strong operator topology. But limn||Tn*e1 || = limn ||en || = 1, hence Tn* dosen't converge to zero in the strong operator topology. So T ® T* is not continuous in the strong operator topology.
(ii) The involution is continuous with respect to the weak operator topology (since | < Tx,y > | = | < T*y,x > |). Hence (i) implies that the weak operator topology and the strong operator topology don't coincide on B(H).
(iii) ||Tn || = ||e1 ||||en || = 1, and by (i) Tn ® 0 in the strong operator topology. Therefore the operator norm is not continuous on B(H).
(iv) Let D = {n[1/2]Tn ; n Î N}. For each neighborhood U(0,x1, ¼,xm,e) of 0 in the strong operator topology, with xk = ån = 1¥ aknen, ||n[1/2]Tnxk|| = n[1/2]|akn|.
But for every 1 £ k £ m, ån = 1¥ |akn |2 < ¥, hence for each e there exists a natural number n such that n[1/2]|akn| < e. So that 0 belongs to the strong closure of D. It follows from the principle of uniform boundedness and ||Ön Tn || = Ön that any sequence in D doesn't converge to 0 in the strong operator topology. Hence the strong operator is not metrizable. Similarly one can show that the weak operator topology is not metrizable.
(v) Let L be the set of all (n,U) in which n Î N and U is a neighborhood of 0 in the strong operator topology on B(H). Then L with the following relation is a directed set:
(m,U) £ (m¢,U¢) Û (m £ m¢  and  U Ê U¢)
Suppose that S is the unilateral shift operator on (en), i.e. S(åk = 1¥ak ek) = åk = 1¥ak ek+1.
Obviously S*(åk = 1¥ak ek) = åk = 1¥ak+1 ek. If l = (ml , Ul ) Î L , limn ||Sn*x || = ml limn (åk = 1¥|ak+n|2)[1/2] ® 0 whenever x = åk = 1¥ak ek Î H. Therefore (ml Sn*)n Î N converges to 0 in the strong operator topology. So that there exists a positive integer number nl such that ml Snl * Î Ul. Set Tl = ml Snl * and Rl = [1/(ml )]Snl . Then liml ||Rl || = liml [1/(ml )] = 0, so that Rl converges to 0 in the norm topology.
If U be a strong neighborhood of 0 and l0 = (1,U), then Tl0 Î Ul0 and for every l ³ l0, Tl Î Ul Í Ul0. therefore (Tl)l Î L converges to 0 in the strong operator topology. But Tl Rl = 1 for all l, hence if the multiplication is jointly continuous in either the weak or the strong operator topology, then 1 = liml Tl Rl = liml Tl liml Rl = 0, a contradiction.
Ref.
[Mur] G.J. Murphy, C*-algebras and operator theory, Academic Press, 1990.


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On 24 Feb 2001, 12:59.